Let $h(x)=\dfrac{\cos(x)}{\ln(x)}$. Find $h'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-x\sin(x)$ (Choice B) B $\dfrac{-x\sin(x)\ln(x)-\cos(x)}{x(\ln(x))^2}$ (Choice C) C $\dfrac{\ln(x)\sin(x)-\cos(x)}{x\ln(x)}$ (Choice D) D $\sin(x)-\dfrac1x$
Solution: $h(x)$ is the quotient of two, more basic, expressions: $\cos(x)$ and $\ln(x)$. Therefore, the derivative of $h$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac{\cos(x)}{\ln(x)}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(\cos(x))\ln(x)-\cos(x)\dfrac{d}{dx}(\ln(x))}{(\ln(x))^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{-\sin(x)\cdot \ln(x)-\cos(x)\cdot\dfrac1x}{(\ln(x))^2}&&\gray{\text{Differentiate }\cos(x)\text{ and }\ln(x)} \\\\ &=\dfrac{-x\sin(x)\ln(x)-\cos(x)}{x(\ln(x))^2}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $h'(x)=\dfrac{-x\sin(x)\ln(x)-\cos(x)}{x(\ln(x))^2}$